= Constant of integration
{title2=$\int f(x)dx=F(x)+C$}
On a connected open <interval>, if $F'=g$, every <antiderivative> of $g$ has the form $F+C$, where $C$ is a <constant function>. Indeed, two solutions satisfy $(G-F)'=0$: the <zero derivative on a connected open set> principle makes their difference constant. In one dimension, the <mean value theorem> proves this on any closed subinterval. https://www.jirka.org/ra/html/sec_mvt.html[Lebl, Basic Analysis I, Proposition 4.2.6: zero derivative on an interval]{ref}
An <initial condition> or <boundary condition> can determine $C$. If a particular <antiderivative> is $F_0$ and $F(x_0)=y_0$, then
$$
\boxed{C=y_0-F_0(x_0).}
$$
For $g(x)=x^2$, $F_0(x)=x^3/3$ and $F(1)=2$, this gives $C=5/3$. Without that reference value, choosing $C=0$ simply selects one member of the family.
The freedom has a physical meaning in a simple kinematic model. A <velocity> $v(t)=at$, with constant <acceleration> $a$, determines <position> only as $x(t)=at^2/2+C$. The <initial condition> $x(0)=x_0$ selects $C=x_0$; changing the starting <position> does not change the prescribed <velocity>.
For the <differential equation> $T'=-B/r^2$ on $r>0$, the <antiderivative> is $T=B/r+C$. The <boundary condition> $T(R_s)=T_s$ gives $C=T_s-B/R_s$. Discarding the <constant of integration> therefore needs a specified condition or an approximation that quantifies its effect. On a domain with several <connected components>, the freedom is larger: <antiderivatives on disconnected domains> require a separate constant on each component.
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